Question

If ${\log }_{7}2=m$, then ${\log }_{49}28$ in terms of $m$ has the value

• A. $\frac{2m+1}{2}$
• B. $4m+2$
• C. $\frac{m+1}{2}$
• D. $2m+2$

Answer 1

The correct option is A $\frac{2m+1}{2}$

As ${\log }_{7}2=m$

${\log }_{49}28={\log }_{7^2}28=\frac{1}{2}\left({\log }_{7}28\right)[\because {\log }_{a^n}{b}^m=\frac{m}{n}{\log }_{a}b]$

$=\frac{1}{2}\left({\log }_{7}7+{\log }_{7}4\right)[\because \log (ab)=\log a+\log b]$

$=\frac{1}{2}(1+2{\log }_{7}2)[\because {\log }_{a}a=1]$

$=\frac{1}{2}(1+2m)=\frac{2m+1}{2}$